Question

我试图找到每个组的一个变量x的唯一值出现次数，由变量/键y定义。

我一直在使用以下代码：

 DT[,length(unique(x)),by=y] -> x_count_per_y

这有效，但有点慢。有没有办法为data.table优化这个，或者这是我应该期待的最快？

Answer 1

使用data.table 1.9.5版本中的uniqueN 它也应该在1.9.4中使用

uniqueN <- function(x) length(attr(data.table:::forderv(x, retGrp=TRUE),"starts",TRUE))

以编程方式使用

byvar = "y"
countvar = "x"
DT[, uniqueN(.SD), by=byvar, .SDcols=countvar]

以下时间：

library(data.table)
library(microbenchmark)
N <- 1e6
DT <- data.table(x = sample(1e5,N,TRUE), y = sample(1e2,N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: milliseconds
#                                         expr      min       lq     mean   median       uq      max neval
#                   DT[, length(unique(x)), y] 85.58602 85.58602 85.58602 85.58602 85.58602 85.58602     1
#                          DT[, uniqueN(x), y] 92.71877 92.71877 92.71877 92.71877 92.71877 92.71877     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"] 97.51024 97.51024 97.51024 97.51024 97.51024 97.51024     1
N <- 1e7
DT <- data.table(x = sample(1e5,N,TRUE), y = sample(1e2,N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: milliseconds
#                                         expr       min        lq      mean    median        uq       max neval
#                   DT[, length(unique(x)), y] 1642.5212 1642.5212 1642.5212 1642.5212 1642.5212 1642.5212     1
#                          DT[, uniqueN(x), y]  843.0670  843.0670  843.0670  843.0670  843.0670  843.0670     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"]  804.7881  804.7881  804.7881  804.7881  804.7881  804.7881     1
N <- 1e7
DT <- data.table(x = sample(1e6,N,TRUE), y = sample(1e5,N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: seconds
#                                         expr      min       lq     mean   median       uq      max neval
#                   DT[, length(unique(x)), y] 3.025365 3.025365 3.025365 3.025365 3.025365 3.025365     1
#                          DT[, uniqueN(x), y] 4.734323 4.734323 4.734323 4.734323 4.734323 4.734323     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"] 5.905721 5.905721 5.905721 5.905721 5.905721 5.905721     1
N <- 1e7
DT <- data.table(x = sample(1e3,N,TRUE), y = sample(1e5,N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: seconds
#                                         expr      min       lq     mean   median       uq      max neval
#                   DT[, length(unique(x)), y] 2.906589 2.906589 2.906589 2.906589 2.906589 2.906589     1
#                          DT[, uniqueN(x), y] 4.731925 4.731925 4.731925 4.731925 4.731925 4.731925     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"] 7.084020 7.084020 7.084020 7.084020 7.084020 7.084020     1
N <- 1e7
DT <- data.table(x = sample(1e6,N,TRUE), y = sample(1e2,N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: milliseconds
#                                         expr      min       lq     mean   median       uq      max neval
#                   DT[, length(unique(x)), y] 1331.244 1331.244 1331.244 1331.244 1331.244 1331.244     1
#                          DT[, uniqueN(x), y]  998.040  998.040  998.040  998.040  998.040  998.040     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"] 1096.867 1096.867 1096.867 1096.867 1096.867 1096.867     1

很大程度上取决于数据，但我已经填写了一个问题来看看这些时间。还有一个角色：

N <- 1e7
DT <- data.table(x = sample(letters,N,TRUE), y = sample(letters[1:10],N,TRUE))
microbenchmark(times=1L,
               DT[, length(unique(x)),y],
               DT[, uniqueN(x),y],
               DT[, uniqueN(.SD), by="y", .SDcols="x"])
# Unit: milliseconds
#                                         expr       min        lq      mean    median        uq       max neval
#                   DT[, length(unique(x)), y] 1304.4865 1304.4865 1304.4865 1304.4865 1304.4865 1304.4865     1
#                          DT[, uniqueN(x), y]  573.8628  573.8628  573.8628  573.8628  573.8628  573.8628     1
#  DT[, uniqueN(.SD), by = "y", .SDcols = "x"]  528.3269  528.3269  528.3269  528.3269  528.3269  528.3269     1

使用data.table优化计算一个变量的唯一值的数量

1 个答案: